Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $z = \dfrac{-a - 1}{a - 7} \div \dfrac{a^2 + 6a + 5}{-3a^2 - 15a} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $z = \dfrac{-a - 1}{a - 7} \times \dfrac{-3a^2 - 15a}{a^2 + 6a + 5} $ First factor the quadratic. $z = \dfrac{-a - 1}{a - 7} \times \dfrac{-3a^2 - 15a}{(a + 1)(a + 5)} $ Then factor out any other terms. $z = \dfrac{-(a + 1)}{a - 7} \times \dfrac{-3a(a + 5)}{(a + 1)(a + 5)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac{ -(a + 1) \times -3a(a + 5) } { (a - 7) \times (a + 1)(a + 5) } $ $z = \dfrac{ 3a(a + 1)(a + 5)}{ (a - 7)(a + 1)(a + 5)} $ Notice that $(a + 5)$ and $(a + 1)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac{ 3a\cancel{(a + 1)}(a + 5)}{ (a - 7)\cancel{(a + 1)}(a + 5)} $ We are dividing by $a + 1$ , so $a + 1 \neq 0$ Therefore, $a \neq -1$ $z = \dfrac{ 3a\cancel{(a + 1)}\cancel{(a + 5)}}{ (a - 7)\cancel{(a + 1)}\cancel{(a + 5)}} $ We are dividing by $a + 5$ , so $a + 5 \neq 0$ Therefore, $a \neq -5$ $z = \dfrac{3a}{a - 7} ; \space a \neq -1 ; \space a \neq -5 $